Chapter 8 Solutions 8.1 (a) P ( V) R becomes 0.0 W (11.6 V) R so R 6.73 Ω (b) V IR so 11.6 V I (6.73 Ω) and I 1.7 A ε IR + Ir so 15.0 V 11.6 V + (1.7 A)r r 1.97 Ω Figure for Goal Solution Goal Solution A battery has an emf of 15.0 V. The terminal voltage of the battery is 11.6 V when it is delivering 0.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance of the battery? G : O : The internal resistance of a battery usually is less than 1 Ω, with physically larger batteries having less resistance due to the larger anode and cathode areas. The voltage of this battery drops significantly (3%), when the load resistance is added, so a sizable amount of current must be drawn from the battery. If we assume that the internal resistance is about 1 Ω, then the current must be about 3 A to give the 3.4 V drop across the battery s internal resistance. If this is true, then the load resistance must be about R 1 V / 3 A 4 Ω. We can find R exactly by using Joule s law for the power delivered to the load resistor when the voltage is 11.6 V. Then we can find the internal resistance of the battery by summing the electric potential differences around the circuit. A : (a) Combining Joule's law, P VI, and the definition of resistance, V IR, gives R V P ( 11.6 V) 0.0 W 6.73 Ω (b) The electromotive force of the battery must equal the voltage drops across the resistances: ε IR + Ir, where I VR. r ε IR I ε ( V)R (15.0 V 11.6 V)(6.73 Ω) 1.97 Ω V 11.6 V L : The resistance of the battery is larger than 1 Ω, but it is reasonable for an old battery or for a battery consisting of several small electric cells in series. The load resistance agrees reasonably well with our prediction, despite the fact that the battery s internal resistance was about twice as large as we assumed. Note that in our initial guess we did not consider the power of the load resistance; however, there is not sufficient information to accurately solve this problem without this data.
Chapter 8 Solutions 135 8. (a) V term IR becomes 10.0 V I (5.60 Ω) so I 1.79 A (b) V term ε Ir becomes 10.0 V ε (1.79 A)(0.00 Ω) so ε 10.4 V 8.3 The total resistance is R 3.00 V 0.600 A 5.00 Ω (a) (b) R lamp R r batteries 5.00 Ω 0.408 Ω 4.59 Ω P batteries P total (0.408 Ω)I 0.0816 8.16% (5.00 Ω)I 8.4 (a) Here ε I(R + r), so I ε R + r 1.6 V (5.00 Ω + 0.0800 Ω).48 A Then, V IR (.48 A) ( 5.00 Ω) 1.4 V (b) Let I 1 and I be the currents flowing through the battery and the headlights, respectively. Then, I 1 I + 35.0 A, and ε I 1 r I R 0 so giving Thus, ε (I + 35.0 A)(0.0800 Ω) + I (5.00 Ω) 1.6 V I 1.93 A V (1.93 A)(5.00 Ω) 9.65 V 8.5 V I 1 R 1 (.00 A)R 1 and V I (R 1 + R ) (1.60 A)(R 1 + 3.00 Ω) Therefore, (.00 A)R 1 (1.60 A)(R 1 + 3.00 Ω) or R 1 1.0 Ω
136 Chapter 8 Solutions 1 8.6 (a) R p ( 17.00 Ω)+ 110.0 Ω ( ) 4.1 Ω R s R 1 + R + R 3 4.00 + 4.1 + 9.00 17.1 Ω (b) V IR 34.0 V I (17.1 Ω) I 1.99 A for 4.00 Ω, 9.00 Ω resistors Applying V IR, (1.99 A)(4.1 Ω) 8.18 V 8.18 V I (7.00 Ω) so I 1.17 A for 7.00 Ω resistor 8.18 V I (10.0 Ω) so I 0.818 A for 10.0 Ω resistor *8.7 If all 3 resistors are placed in parallel, 1 R 1 500 + 50 5 500 and R 100 Ω *8.8 For the bulb in use as intended, I P V 75.0 W V 0.65 A and R 10 V I 10 V 0.65 A 19 Ω Now, presuming the bulb resistance is unchanged, I 10 V 193.6 Ω 0.60 A Across the bulb is V IR 19 Ω(0.60 A) 119 V so its power is P ( V)I 119 V(0.60 A) 73.8 W
Chapter 8 Solutions 137 8.9 If we turn the given diagram on its side, we find that it is the same as Figure (a). The 0.0-Ω and 5.00-Ω resistors are in series, so the first reduction is as shown in (b). In addition, since the 10.0-Ω, 5.00-Ω, and 5.0-Ω resistors are then in parallel, we can solve for their equivalent resistance as: R eq 1 ( ).94 Ω 1 + 1 + 1 10.0 Ω 5.00 Ω 5.0 Ω This is shown in Figure (c), which in turn reduces to the circuit shown in (d). Next, we work backwards through the diagrams applying I V/R and V IR. The 1.94-Ω resistor is connected across 5.0-V, so the current through the battery in every diagram is I V R 5.0 V 1.94 Ω 1.93 A (a) In Figure (c), this 1.93 A goes through the.94-ω equivalent resistor to give a potential difference of: V IR (1.93 A)(.94 Ω) 5.68 V From Figure (b), we see that this potential difference is the same across V ab, the 10-Ω resistor, and the 5.00-Ω resistor. (b) (b) Therefore, V ab 5.68 V (a) Since the current through the 0.0-Ω resistor is also the current through the 5.0-Ω line ab, I V ab R ab 5.68 V 5.0 Ω 0.7 A 7 ma (c) (d) ( 10 V) 8.10 10 V IR eq I ρl + ρl + ρl + ρl, or Iρl A 1 A A 3 A 4 1 + 1 + 1 + 1 A 1 A A 3 A 4 V Iρl A ( 10 V) 1 A + 1 + 1 + 1 9.5 V A 1 A A 3 A 4
138 Chapter 8 Solutions 8.11 (a) Since all the current flowing in the circuit must pass through the series 100-Ω resistor, P RI P max RI max so I max P R 5.0 W 100 Ω 0.500 A R eq 100 Ω + 1 100 + 1 1 100 Ω 150 Ω V max R eq I max 75.0 V (b) P ( V)I (75.0 V)(0.500 A) 37.5 W total power P 1 5.0 W P P 3 RI (100 Ω)(0.50 A) 6.5 W 8.1 Using.00-Ω, 3.00-Ω, 4.00-Ω resistors, there are 7 series, 4 parallel, and 6 mixed combinations: Series Parallel Mixed The resistors may be arranged in patterns:.00 Ω 6.00 Ω 3.00 Ω 7.00 Ω 4.00 Ω 9.00 Ω 5.00 Ω 0.93 Ω1.56 Ω 1.0 Ω.00 Ω 1.33 Ω. Ω 1.71 Ω 3.71 Ω 4.33 Ω 5.0 Ω 8.13 The potential difference is the same across either combination. V IR 3I 1 + R 500 1 ( ) 1 R + 500 1 so R 1 R + 1 500 3 3 and R 1000 Ω 1.00 kω 8.14 If the switch is open, I ε /( R + R) and P ε R /( R + R) If the switch is closed, I ε /(R + R / ) and P ε ( R /)/(R + R /) Then, ε R ε ( R + R) R (R + R /) R +R R + R / R +R R + R The condition becomes R R / so R R (1.00 Ω) 1.41 Ω
Chapter 8 Solutions 139 1 8.15 R p 3.00 + 1 1.00 1 0.750 Ω R s (.00 + 0.750 + 4.00) Ω6.75 Ω I battery V R s 18.0 V 6.75 Ω.67 A P I R: P (.67 A) (.00 Ω) P 14. W in.00 Ω P 4 (.67 A) ( 4.00 Ω) 8.4 W in 4.00 Ω V (.67 A) (.00 Ω) 5.33 V, V 4 (.67 A) ( 4.00 Ω) 10.67 V V p 18.0 V V V 4.00 V ( V 3 V 1 ) ( P 3 V 3) R 3 ( P 1 V 1) R 1 (.00 V) 1.33 W in 3.00 Ω 3.00 Ω (.00 V) 4.00 W in 1.00 Ω 1.00 Ω 8.16 Denoting the two resistors as x and y, x + y 690, and 1 150 1 x + 1 y 1 150 1 x + 1 690 x (690 x) + x x(690 x) x 690x + 103,500 0 x 690 ± (690) 414,000 x 470 Ω y 0 Ω
140 Chapter 8 Solutions 8.17 (a) V IR: 33.0 V I 1 ( 11.0 Ω) 33.0 V I (.0 Ω) I 1 3.00 A I 1.50 A P I R: P 1 ( 3.00 A) ( 11.0 Ω) P ( 1.50 A) (.0 Ω) P 1 99.0 W P 49.5 W The 11.0-Ω resistor uses more power. (b) P 1 + P 148 W P I( V) ( 4.50) ( 33.0) 148 W (c) R s R 1 + R 11.0 Ω+.0 Ω33.0 Ω V IR: 33.0 V I( 33.0 Ω), so I 1.00 A P I R: P 1 ( 1.00 A) ( 11.0 Ω) P ( 1.00 A) (.0 Ω) P 1 11.0 W P.0 W The.0-Ω resistor uses more power. (d) P 1 + P I ( R 1 + R ) ( 1.00 A) ( 33.0 Ω) 33.0 W P I( V) ( 1.00 A) ( 33.0 V) 33.0 W (e) The parallel configuration uses more power. 8.18 +15.0 (7.00)I 1 (.00)(5.00) 0 5.00 7.00I 1 so I 1 0.714 A I 3 I 1 + I.00 A 0.714 + I.00 so I 1.9 A +ε.00(1.9) (5.00)(.00) 0 ε 1.6 V
Chapter 8 Solutions 141 8.19 We name the currents I 1, I, and I 3 as shown. From Kirchhoff's current rule, I 3 I 1 + I Applying Kirchhoff's voltage rule to the loop containing I and I 3, 1.0 V (4.00)I 3 (6.00)I 4.00 V 0 8.00 (4.00)I 3 + (6.00)I Applying Kirchhoff's voltage rule to the loop containing I 1 and I, (6.00)I 4.00 V + (8.00)I 1 0 (8.00)I 1 4.00 + (6.00)I Solving the above linear systems, I 1 846 ma, I 46 ma, I 3 1.31 A All currents flow in the directions indicated by the arrows in the circuit diagram. *8.0 The solution figure is shown to the right. *8.1 We use the results of Problem 19. (a) By the 4.00-V battery: U ( V)It 4.00 V( 0.46 A)10 s J By the 1.0-V battery: 1.0 V (1.31 A) 10 s 1.88 kj (b) By the 8.00 Ω resistor: I Rt (0.846 A) (8.00 Ω ) 10 s 687 J By the 5.00 Ω resistor: (0.46 A) (5.00 Ω ) 10 s 18 J By the 1.00 Ω resistor: (0.46 A) (1.00 Ω ) 10 s 5.6 J By the 3.00 Ω resistor: (1.31 A) (3.00 Ω ) 10 s 616 J By the 1.00 Ω r e s i s t o r : (1.31 A) (1.00 Ω ) 10 s 05 J (c) J + 1.88 kj 1.66 kj from chemical to electrical. 687 J + 18 J + 5.6 J + 616 J + 05 J 1.66 kj from electrical to heat.
14 Chapter 8 Solutions 8. We name the currents I 1, I, and I 3 as shown. [1] 70.0 60.0 I (3.00 kω) I 1 (.00 kω) 0 [] 80.0 I 3 (4.00 kω) 60.0 I (3.00 kω) 0 [3] I I 1 + I 3 (a) Substituting for I and solving the resulting simultaneous equations yields I 1 0.385 ma (through R 1 ) I 3.69 ma (through R 3 ) I 3.08 ma (through R ) (b) V cf 60.0 V (3.08 ma)(3.00 kω) 69. V Point c is at higher potential. 8.3 Label the currents in the branches as shown in the first figure. Reduce the circuit by combining the two parallel resistors as shown in the second figure. Apply Kirchhoff s loop rule to both loops in Figure (b) to obtain: (.71R)I 1 + ( 1.71R)I 50 and ( 1.71R)I 1 + ( 3.71R)I 500 (a) With R 1000 Ω, simultaneous solution of these equations yields: I 1 10.0 ma and I 130.0 ma From Figure (b), V c V a ( I 1 + I )( 1.71R) 40 V Thus, from Figure (a), I 4 V c V a 4R 40 V 60.0 ma 4000 Ω (b) Finally, applying Kirchhoff s point rule at point a in Figure (a) gives: I I 4 I 1 60.0 ma 10.0 ma +50.0 ma, or I 50.0 ma flowing from point a to point e.
Chapter 8 Solutions 143 8.4 Name the currents as shown in the figure to the right. Then w + x + z y. Loop equations are 00w 40.0 + 80.0x 0 80.0x + 40.0 + 360 0.0y 0 + 360 0.0y 70.0z + 80.0 0 Eliminate y by substitution. Eliminate x : x.50 w + 0.500 400 100x 0.0w 0.0 z 0 440 0.0w 0.0x 90.0 z 0 350 70w 0.0 z 0 430 70.0w 90.0 z 0 Eliminate z 17.5 13.5w to obtain 430 70.0w 1575 + 115 w 0 w 70.0/70.0 1.00 A upward in 00 Ω Now z 4.00 A upward in 70.0 Ω x 3.00 A upward in 80.0 Ω y 8.00 A downward in 0.0 Ω and for the 00 Ω, V IR (1.00 A)(00 Ω) 00 V 8.5 Using Kirchhoff s rules, 1.0 ( 0.0100)I 1 ( 0.0600)I 3 0 10.0 + ( 1.00)I ( 0.0600)I 3 0 and I 1 I + I 3 1.0 ( 0.0100)I ( 0.0700)I 3 0 10.0 + ( 1.00)I ( 0.0600)I 3 0 Solving simultaneously, dead battery, I 0.83 A downward in the and I 3 171 A downward in the starter.
144 Chapter 8 Solutions 8.6 V ab ( 1.00)I 1 + ( 1.00) ( I 1 I ) V ab ( 1.00)I 1 + ( 1.00)I + ( 5.00) ( I I 1 + I ) V ab ( 3.00) ( I I 1 )+ ( 5.00) ( I I 1 + I ) Let I 1.00 A, I 1 x, and I y Then, the three equations become: V ab.00 x y, or y.00x V ab V ab 4.00 x + 6.00 y + 5.00 and V ab 8.00 8.00 x + 5.00 y Substituting the first into the last two gives: 7.00V ab 8.00 x + 5.00 and 6.00V ab.00 x + 8.00 Solving these simultaneously yields V ab 7 17 V Then, R ab V ab I 7 17 V 1.00 A or R ab 7 17 Ω 8.7 We name the currents I 1, I, and I 3 as shown. (a) I 1 I + I 3 Counterclockwise around the top loop, 1.0 V (.00 Ω)I 3 (4.00 Ω)I 1 0 Traversing the bottom loop, 8.00 V (6.00 Ω)I + (.00 Ω)I 3 0 I 1 3.00 1 I 3 I 4 3 + 1 3 I 3 and I 3 909 ma (b) V a (0.909 A)(.00 Ω) V b V b V a 1.8 V
Chapter 8 Solutions 145 8.8 We apply Kirchhoff's rules to the second diagram. 50.0.00I 1.00I 0 (1) 0.0.00I 3 +.00I 0 () I 1 I + I 3 (3) Substitute (3) into (1), and solve for I 1, I, and I 3 I 1 0.0 A; I 5.00 A; I 3 15.0 A Then apply P I R to each resistor: (.00 Ω) 1 : P I 1 (.00 Ω) (0.0 A) (.00 Ω) 800 W (4.00 Ω) : P 5.00 A (4.00 Ω) 5.0 W (Half of I goes through each) (.00 Ω) 3 : P I 3 (.00 Ω) (15.0 A) (.00 Ω) 450 W 8.9 (a) RC (1.00 10 6 Ω)(5.00 10 6 F) 5.00 s (b) Q Cε (5.00 10 6 C)(30.0 V) 150 µc ε 30.0 (c) I(t) R e t/rc 1.00 10 6 exp 10.0 (1.00 10 6 )(5.00 10 6 ) 4.06 µa 8.30 (a) I(t) I 0 e t/rc I 0 Q RC 5.10 10 6 C (1300 Ω)(.00 10 9 F) 1.96 A I(t) (1.96 A) exp 9.00 10 6 s (1300 Ω)(.00 10 9 F) 61.6 ma (b) q(t) Qe t/rc (5.10 µc) exp 8.00 10 6 s (1300 Ω)(.00 10 9 F) 0.35 µc (c) The magnitude of the current is I 0 1.96 A
146 Chapter 8 Solutions 8.31 U 1 C( V) and V QC Therefore, U Q C and when the charge decreases to half its original value, the stored energy is one-quarter its original value: U f 1 4 U 0 8.3 (a) τ RC (1.50 10 5 Ω)(10.0 10 6 F) 1.50 s (b) τ (1.00 10 5 Ω)(10.0 10 6 F) 1.00 s (c) The battery carries current 10.0 V 50.0 10 3 Ω 00 µa The 100 kω carries current of magnitude I I 0 e t/rc 10.0 V 100 10 3 Ω e t/ 1.00 s So the switch carries downward current t / 1.00 s 00 µa + (100 µa)e 8.33 (a) Call the potential at the left junction V L and at the right V R. After a "long" time, the capacitor is fully charged. V L 8.00 V because of voltage divider: I L 10.0 V 5.00 Ω.00 A V L 10.0 V (.00 A)(1.00 Ω) 8.00 V Likewise, V R.00 Ω.00 Ω + 8.00 Ω 10.0 V.00 V or I R 10.0 V 10.0 Ω 1.00 A V R (10.0 V) (8.00 Ω)(1.00 A).00 V Therefore, V V L V R 8.00.00 6.00 V 1 (b) Redraw the circuit R ( 1/ 9.00 Ω)+ 1/ 6.00 Ω RC 3.60 10 6 s ( ) 3.60 Ω and e t/rc 1 10 so t RC ln 10 8.9 µs
Chapter 8 Solutions 147 8.34 (a) τ RC ( 4.00 10 6 Ω) ( 3.00 10 6 F) 1.0 s ε (b) I R e t/rc 1.0 4.00 10 6 e t/1.0 s q Cε [ 1 e t/rc ] 3.00 10 6 ( 1.0) [ 1 e t/1.0 ) q 36.0 µc[ 1 e t/1.0 ] I 3.00 µae t/1.0 8.35 V 0 Q C Then, if q(t) Qe t/rc V(t) V 0 e t/rc V() t e t/rc V 0 Therefore 1 exp 4.00 R 3.60 10 6 ( ) ln 1 4.00 R 3.60 10 6 R 1.60 MΩ ( ) 8.36 V 0 Q C Then, if q(t) Qe trc V(t) ( V 0 )e trc and V(t) RC e t ( V 0 ) When V(t) 1 ( V 0), then e trc 1 Thus, R t RC ln 1 ln t C ln ( )
148 Chapter 8 Solutions 8.37 q(t) Q[ 1 e t/rc ] so q(t) Q 1 e t/rc 0.600 1 e 0.900/RC or e 0.900/RC 1 0.600 0.400 0. 900 ln( 0. 400) RC thus RC 0. 900 ln( 0. 400) 0.98 s 8.38 Applying Kirchhoff s loop rule, I g ( 75.0 Ω)+ ( I I g )R p 0 Therefore, if I 1.00 A when I g 1.50 ma, R p I g ( 75.0 Ω) ( I I g ) ( ) 75.0 Ω ( ) 1. 50 10 3 A 1.00 A 1. 50 10 3 A 0.113 Ω 8.39 Series Resistor Voltmeter V IR: 5.0 1.50 10-3 (R s + 75.0) Solving, R s 16.6 kω Figure for Goal Solution Goal Solution The galvanometer described in the preceding problem can be used to measure voltages. In this case a large resistor is wired in series with the galvanometer in a way similar to that shown in Figure P8.4b This arrangement, in effect, limits the current that flows through the galvanometer when large voltages are applied. Most of the potential drop occurs across the resistor placed in series. Calculate the value of the resistor that enables the galvanometer to measure an applied voltage of 5.0 V at full-scale deflection. G : O : The problem states that the value of the resistor must be large in order to limit the current through the galvanometer, so we should expect a resistance of kω to MΩ. The unknown resistance can be found by applying the definition of resistance to the portion of the circuit shown in Figure 8.4b. A : V ab 5.0 V; From Problem 38, I 1.50 ma and R g 75.0 Ω. For the two resistors in series, R eq R s + R g so the definition of resistance gives us: V ab I(R s + R g ) Therefore, R s V ab I 5.0 V R g 1. 50 10 3 75.0 Ω16.6 kω A L : The resistance is relatively large, as expected. It is important to note that some caution would be necessary if this arrangement were used to measure the voltage across a circuit with a comparable resistance. For example, if the circuit resistance was 17 kω, the voltmeter in this problem would cause a measurement inaccuracy of about 50%, because the meter would divert about half the current that normally would go through the resistor being measured. Problems 46 and 59 address a similar concern about measurement error when using electrical meters.
Chapter 8 Solutions 149 8.40 We will use the values required for the 1.00-V voltmeter to obtain the internal resistance of the galvanometer. V I g (R + r g ) Solve for r g : r g V I g R 1.00 V 1.00 10-3 A 900 Ω 100 Ω We then obtain the series resistance required for the 50.0-V voltmeter: R V I g r g 50.0 V 1.00 10-3 A 100 Ω 49.9 kω 8.41 V I g r g ( I I g )R p, or R p I g r g I I g ( ) ( ) I g 60.0 Ω ( I I g ) Therefore, to have I 0.100 A 100 ma when I g 0.500 ma: 0.500 ma R p ( ) 60.0 Ω 99.5 ma ( ) 0.30 Ω Figure for Goal Solution Goal Solution Assume that a galvanometer has an internal resistance of 60.0 Ω and requires a current of 0.500 ma to produce full-scale deflection. What resistance must be connected in parallel with the galvanometer if the combination is to serve as an ammeter that has a full-scale deflection for a current of 0.100 A? G : O : A : An ammeter reads the flow of current in a portion of a circuit; therefore it must have a low resistance so that it does not significantly alter the current that would exist without the meter. Therefore, the resistance required is probably less than 1 Ω. From the values given for a full-scale reading, we can find the voltage across and the current through the shunt (parallel) resistor, and the resistance value can then be found from the definition of resistance. The voltage across the galvanometer must be the same as the voltage across the shunt resistor in parallel, so when the ammeter reads full scale, V ( 0.500 ma) ( 60.0 Ω) 30.0 mv Through the shunt resistor, I 100 ma 0.500 ma 99.5 ma Therefore, R V I 30.0 mv 99.5 ma 0.30 Ω L : The shunt resistance is less than 1 Ω as expected. It is important to note that some caution would be necessary if this meter were used in a circuit that had a low resistance. For example, if the circuit resistance was 3 Ω, adding the ammeter to the circuit would reduce the current by about 10%, so the current displayed by the meter would be lower than without the meter. Problems 46 and 59 address a similar concern about measurement error when using electrical meters.
150 Chapter 8 Solutions 8.4 R x R R 3 R 1 R R 3.50 R 1000 Ω.50 400 Ω 8.43 Using Kirchhoff s rules with R g << 1, ( 1.0 Ω)I 1 + ( 14.0 Ω)I 0, so I 1 3 I 70.0 1.0I 1 7.00( I 1 + I g ) 0, and 70.0 14.0I 7.00( I I g ) 0 The last two equations simplify to 1.0 Ω I 1 I 14.0 Ω + G I g 70.0 V I - I g 7.00 Ω I 1 + I g 7.00 Ω 10.0 4.00( 3 I ) I g, and 10.0 3.00I I g Solving simultaneously yields: I g 0.588 A 8.44 R ρl A and R i ρl i A i But, V AL A i L i, so R ρl V and R i ρl i V Therefore, R ρ ( L i + L) V [ ( )] ρl i 1 + LL i V R i [ 1 + α] where α L L This may be written as: R R i (1 + α + α ) 8.45 ε x R s ε s R s ; ε x ε sr x 48.0 Ω R s 36.0 Ω (1.0186 V) 1.36 V
Chapter 8 Solutions 151 *8.46 (a) In Figure (a), the emf sees an equivalent resistance of 00.00 Ω. I 6.000 0 V 0.030 000 A 00.00 Ω The terminal potential difference is 6.0000 V 0.000 Ω V 0.000 Ω 0.000 Ω 180.00 Ω 180.00 Ω 180.00 Ω (a) (b) (c) V IR ( 0.030 000 A) ( 180.00 Ω) 5.400 0 V A V A 1 (b) In Figure (b), R eq 180.00 Ω + 1 0 000 Ω 1 178.39 Ω (c) The equivalent resistance across the emf is The ammeter reads and the voltmeter reads In Figure (c), Therefore, the emf sends current through The current through the battery is but not all of this goes through the ammeter. The voltmeter reads 178.39 Ω+0.500 00 Ω+0.000 Ω198.89 Ω I ε R 6.000 0 V 198.89 Ω 0.030 167 A V IR ( 0.030 167 A) ( 178.39 Ω) 5.381 6 V 1 180.50 Ω + 1 0 000 Ω 178.89 Ω R tot 178.89 Ω+0.000 Ω198.89 Ω I 6.000 0 V 198.89 Ω 1 0.030 168 A V IR ( 0.030 168 A) ( 178.89 Ω) 5.396 6 V The ammeter measures current I V R 5.396 6 V 0.09 898 A 180.50 Ω The connection shown in Figure (c) is better than that shown in Figure (b) for accurate readings. 8.47 (a) P I( V) So for the Heater, I P V 1500 W 10 V 1.5 A For the Toaster, And for the Grill, I 750 W 10 W 6.5 A I 1000 W 10 V 8.33 A (Grill) (b) 1.5 + 6.5 + 8.33 7.1 A sufficient. The current draw is greater than 5.0 amps, so this would not be
15 Chapter 8 Solutions 8.48 (a) P I R I ρl A (1.00 A) (1.70 10 8 Ω m)(16.0 ft)(0.3048 m / ft) π(0.51 10 3 m) 0.101 W (b) P I R 100(0.101 Ω) 10.1 W 8.49 I Al R Al I Cu R Cu so I Al R Cu I Cu R Al ρ Cu I Cu 1.70 ( ρ Al.8 0.0 ) 0.776(0.0) 15.5 A *8.50 (a) Suppose that the insulation between either of your fingers and the conductor adjacent is a chunk of rubber with contact area 4 mm and thickness 1 mm. Its resistance is ( )( 10 3 m) R ρl A 10 13 Ω m 4 10 6 m 10 15 Ω The current will be driven by 10 V through total resistance (series) 10 15 Ω+10 4 Ω+ 10 15 Ω 5 10 15 Ω It is: I V R ~ 10 V 5 10 15 Ω ~10 14 A (b) The resistors form a voltage divider, with the center of your hand at potential V h, where V h is the potential of the "hot" wire. The potential difference between your finger and thumb is V IR ~ ( 10 14 A) ( 10 4 Ω)~10 10 V. So the points where the rubber meets your fingers are at potentials of ~ V h + 10 10 V and ~ V h 10 10 V *8.51 The set of four batteries boosts the electric potential of each bit of charge that goes through them by 4 1.50 V 6.00 V. The chemical energy they store is U q V (40 C)(6.00 J/C) 1440 J The radio draws current I V R 6.00 V 00 Ω 0.0300 A So, its power is P ( V)I (6.00 V)(0.0300 A) 0.180 W 0.180 J/s Then for the time the energy lasts, we have P Et: t E P 1440 J 0.180 J / s 8.00 103 s We could also compute this from I Q/t: t Q I 40 C 0.0300 A 8.00 103 s. h
Chapter 8 Solutions 153 *8.5 I ε R + r, so P I R ε R ( R + r) or R + r P R ( ) ε Let x ε P, then ( R + r) xr or R + ( r x)r r 0 With r 1.0 Ω, this becomes which has solutions of R + (.40 x)r 1. 44 0, R (.40 x )± (.40 x ) 5.76 + 4.1 ± 4.1 (a) With ε 90. V and P 1.8 W, x 6.61: R ( ) 5.76 3.84 Ω or 0.375 Ω (b) For ε ε 90. V and P 1. W, x P 3.99 R +1. 59 ± ( 1. 59) 5.76 1. 59 ± 3. The equation for the load resistance yields a complex number, so there is no resistance that will extract 1. W from this battery. The maximum power output occurs when R r 1.0 Ω, and that maximum is: P max ε 4r 17.6 W 8.53 Using Kirchhoff s loop rule for the closed loop, +1.0.00I 4.00 I 0, so I.00 A V b V a +4.00 V (.00 A) ( 4.00 Ω) ( 0) ( 10.0 Ω) 4.00 V Thus, V ab 4.00 V and point a is at the higher potential. 8.54 The potential difference across the capacitor V() V t max [ 1 e trc ] Using 1 Farad 1 s Ω, 4.00 V 10.0 V ( ) 1 e 3.00 s Therefore, 0.400 1.00 e ( 3.00 105 Ω) R or e ( 3.00 105 Ω) R 0.600 ( ) ( ) R 10.0 10 6 s Ω Taking the natural logarithm of both sides, 3.00 105 Ω R ln( 0.600) and R 3.00 105 Ω ln 0.600 ( ) +5.87 10 5 Ω 587 kω
154 Chapter 8 Solutions 8.55 Let the two resistances be x and y. x y Then, R s x + y P s I and R p xy x + y P p I 5 W 9.00 Ω y 9.00 Ω x 5.00 A ( ) 50.0 W 5.00 A ( ).00 Ω x y so ( ) ( ).00 Ω x 9.00 x + 18.0 0 x 9.00 Ω x x + 9.00 Ω x Factoring the second equation, ( x 6.00) ( x 3.00) 0 so x 6.00 Ω or x 3.00 Ω Then, y 9.00 Ω x gives y 3.00 Ω or y 6.00 Ω The two resistances are found to be 6.00 Ω and 3.00 Ω. 8.56 Let the two resistances be x and y. Then, R s x + y P s I and R p xy x + y P p I. From the first equation, y P s x, and the second I becomes x( P s I x) x + ( P s I x) P p I or x P s I x + P s P p I 4 0. Using the quadratic formula, x P s ± P s 4P s P p I. Then, y P s I x gives y P s > P s 4P s P p I. The two resistances are P s + P s 4P s P p I and P s P s 4P s P p I
Chapter 8 Solutions 155 8.57 The current in the simple loop circuit will be I ε R + r (a) V ter ε Ir (b) I ε R + r εr R + r (c) P I R ε R (R + r) and and V ter ε as R I ε r as R 0 dp dr ε R( )(R + r) 3 + ε ε (R + r) R ε (R + r) 3 + (R + r) 0 Figure for Goal Solution Then R R + r and R r Goal Solution A battery has an emf ε and internal resistance r. A variable resistor R is connected across the terminals of the battery. Determine the value of R such that (a) the potential difference across the terminals is a maximum, (b) the current in the circuit is a maximum, (c) the power delivered to the resistor is a maximum. G : O : If we consider the limiting cases, we can imagine that the potential across the battery will be a maximum when R (open circuit), the current will be a maximum when R 0 (short circuit), and the power will be a maximum when R is somewhere between these two extremes, perhaps when R r. We can use the definition of resistance to find the voltage and current as functions of R, and the power equation can be differentiated with respect to R. A : (a) The battery has a voltage V terminal ε Ir εr R + r (b) The circuit's current is I ε R + r or as R, V terminal ε or as R 0, I ε r (c) The power delivered is P I R ε R (R + r) To maximize the power P as a function of R, we differentiate with respect to R, and require that dp /dr 0 Then R R + r and R r dp dr ε R( )(R + r) 3 + ε (R + r) ε R (R + r) 3 + ε (R + r) 0 L : The results agree with our predictions. Making load resistance equal to the source resistance to maximize power transfer is called impedance matching.
156 Chapter 8 Solutions 8.58 (a) ε I(ΣR) (ε 1 + ε ) 0 40.0 V (4.00 A) [(.00 + 0.300 + 0.300 + R)Ω] (6.00 + 6.00) V 0; so R 4.40 Ω (b) Inside the supply, Inside both batteries together, For the limiting resistor, P I R ( 4.00 A) (.00 Ω) 3.0 W P I R ( 4.00 A) ( 0.600 Ω) 9.60 W P ( 4.00 A) ( 4.40 Ω) 70.4 W (c) P I(ε 1 + ε ) (4.00 A) [(6.00 + 6.00)V] 48.0 W 8.59 Let R m measured value, R actual value, I R current through the resistor R (a) I current measured by the ammeter. (a) When using circuit (a), I R R V 0 000(I I R ) or R 0 000 I 1 I R But since I V R m and I R V R, we have I I R R R m (b) Figure for Goal solution and R 0 000 (R R m) R m (1) When R > R m, we require (R R m ) R 0.0500 Therefore, R m R (1 0.0500) and from (1) we find (b) When using circuit (b), R 1050 Ω I R R V I R (0.5 Ω). But since I R V R m, R m (0.500 + R) () When R m > R, we require (R m R) R 0.0500 From () we find R 10.0 Ω
Chapter 8 Solutions 157 Goal Solution The value of a resistor R is to be determined using the ammeter-voltmeter setup shown in Figure P8.59. The ammeter has a resistance of 0.500 Ω, and the voltmeter has a resistance of 0000 Ω. Within what range of actual values of R will the measured values be correct to within 5.00% if the measurement is made using (a) the circuit shown in Figure P8.59a (b) the circuit shown in Figure P8.59b? G : An ideal ammeter has zero resistance, and an ideal voltmeter has infinite resistance, so that adding the meter does not alter the current or voltage of the existing circuit. For the non-ideal meters in this problem, a low values of R will give a large voltage measurement error in circuit (b), while a large value of R will give significant current measurement error in circuit (a). We could hope that these meters yield accurate measurements in either circuit for typical resistance values of 1 Ω to 1 MΩ. O : The definition of resistance can be applied to each circuit to find the minimum and maximum current and voltage allowed within the 5.00% tolerance range. A : (a) In Figure P8.59a, at least a little current goes through the voltmeter, so less current flows through the resistor than the ammeter reports, and the resistance computed by dividing the voltage by the inflated ammeter reading will be too small. Thus, we require that V/ I 0.950R where I is the current through the ammeter. Call I R the current through the resistor; then I I R is the current in the voltmeter. Since the resistor and the voltmeter are in parallel, the voltage across the meter equals the voltage across the resistor. Applying the definition of resistance: Our requirement is V I R R ( I I R )( 0000 Ω) so I I R (R + 0000 Ω) I R R 0.95R I R (R + 0000 Ω) 0000 Ω 0000 Ω Solving, 0000 Ω 0.95(R + 0000 Ω)0.95R + 19000 Ω and R 1000 Ω or R 1.05 kω 0.95 (b) If R is too small, the resistance of an ammeter in series will significantly reduce the current that would otherwise flow through R. In Figure 8.59b, the voltmeter reading is I( 0.500 Ω)+ IR, at least a little larger than the voltage across the resistor. So the resistance computed by dividing the inflated voltmeter reading by the ammeter reading will be too large. We require V I 1.05R so that I (0.500 Ω) + IR 1.05R I Thus, 0.500 Ω 0.0500R and R 10.0 Ω L : The range of R values seems correct since the ammeter s resistance should be less than 5% of the smallest R value ( 0.500 Ω 0.05R means that R should be greater than 10 Ω), and R should be less than 5% of the voltmeter s internal resistance ( R 0.05 0 kω1 kω ). Only for the restricted range between 10 ohms and 1000 ohms can we indifferently use either of the connections (a) and (b) for a reasonably accurate resistance measurement. For low values of the resistance R, circuit (a) must be used. Only circuit (b) can accurately measure a large value of R.
158 Chapter 8 Solutions 8.60 The battery supplies energy at a changing rate Then the total energy put out by the battery is de dt P EI E E R e 1/RC de t0 ε R exp t RC dt ε de R ( RC) exp t 0 RC dt ε C exp t RC RC 0 ε C[0 1] ε C The heating power of the resistor is So the total heat is de ε dt P V R I I t R R exp R RC de 0 ε R exp t RC dt ε de R RC 0 exp t RC dt ε C t exp RC RC 0 ε C ε [0 1] C The energy finally stored in the capacitor is U 1 C ( V) 1 C ε. Thus, energy is conserved: ε C 1 ε C + 1 ε C and resistor and capacitor share equally in the energy from the battery. 8.61 (a) q C( V) [ 1 e trc ] q ( 1.00 10 6 F) ( 10.0 V) 9.93 µc (b) I dq dt V R RC e t 10.0 V I.00 10 6 Ω e 5.00 3.37 10 8 A 33.7 na (c) du dt d 1 q dt C q dq C dt q C I du dt 9.93 10 6 C 1.00 10 6 CV 3.37 10 8 A ( ) 3.34 10 7 W 334 nw (d) P battery IE ( 3.37 10 8 A) ( 10.0 V) 3.37 10 7 W 337 nw
Chapter 8 Solutions 159 8.6 Start at the point when the voltage has just reached V and the switch has just closed. The voltage is V 3 3 and is decaying towards 0 V with a time constant R B C. V C (t) 3 V e t/r B C We want to know when V C (t) will reach 1 3 V. Therefore, or t 1 R B C ln 1 3 V 3 V e t/rbc or e t/rbc 1 After the switch opens, the voltage is 1 V, increasing toward V with time constant ( R 3 A + R B)C: V C (t) V 3 V e t/(r A +R B )C When V C (t) 3 V, 3 V V 3 Ve t/(r A +R B )C or e t/(r A +R B )C 1 so t (R A + R B )C ln and T t 1 + t (R A + R B )C ln 8.63 (a) First determine the resistance of each light bulb: P ( V) R R ( V) P (10 V) 60.0 W 40 Ω We obtain the equivalent resistance R eq of the network of light bulbs by applying Equations 8.6 and 8.7: 1 R eq R 1 + 40 Ω+10 Ω360 Ω ( 1/R )+ ( 1/R 3 ) The total power dissipated in the 360 Ω is P ( V) (10 V) R eq 360 Ω 40.0 W (b) The current through the network is given by P I R eq : I P R eq 40.0 W 360 Ω 1 3 A The potential difference across R 1 is V 1 IR 1 1 3 A (40 Ω) 80.0 V The potential difference V 3 across the parallel combination of R and R 3 is V 3 IR 3 1 3 A 1 ( 1/ 40 Ω)+ ( 1/ 40 Ω ) 40.0 V
160 Chapter 8 Solutions 8.64 V IR (a) 0.0 V (1.00 10-3 A)(R 1 + 60.0 Ω) R 1 1.994 10 4 Ω 19.94 kω (b) 50.0 V (1.00 10-3 A)(R + R 1 + 60.0 Ω) R 30.0 kω (c) 100 V (1.00 10-3 A)(R 3 + R 1 + 60.0 Ω) R 3 50.0 kω 8.65 Consider the circuit diagram shown, realizing that I g 1.00 ma. For the 5.0 ma scale: ( 4.0 ma) ( R 1 + R + R 3 ) ( 1.00 ma) ( 5.0 Ω) or R 1 + R + R 3 5.0 4.0 Ω (1) For the 50.0 ma scale: ( 49.0 ma) ( R 1 + R ) ( 1.00 ma) ( 5.0 Ω+R 3 ) or 49.0( R 1 + R ) 5.0 Ω+R 3 () For the 100 ma scale: ( 99.0 ma)r 1 ( 1.00 ma) ( 5.0 Ω+R + R 3 ) or 99.0R 1 5.0 Ω+R + R 3 (3) Solving (1), (), and (3) simultaneously yields R 1 0.60 Ω, R 0.61 Ω, R 3 0.51 Ω 8.66 Ammeter: I g r ( 0.500 A I g )( 0.0 Ω) or Voltmeter: I g ( r + 0.0 Ω) 0.110 V (1).00 V I g ( r + 500 Ω) () Solve (1) and () simultaneously to find: I g 0.756 ma and r 145 Ω
Chapter 8 Solutions 161 8.67 (a) After steady-state conditions have been reached, there is no DC current through the capacitor. Thus, for R 3 : I R3 0 (steady-state) For the other two resistors, the steady-state current is simply determined by the 9.00-V emf across the 1-k Ω and 15-k Ω resistors in series: ε For R 1 and R : I (R1 +R ) 9.00 V 333 µa (steady-state) R 1 + R (1.0 kω+15.0 kω) (b) After the transient currents have ceased, the potential difference across C is the same as the potential difference across R ( IR ) because there is no voltage drop across R 3. Therefore, the charge Q on C is Q C ( V) R C (IR ) (10.0 µf)(333 µa)(15.0 k Ω) 50.0 µc (c) When the switch is opened, the branch containing R 1 is no longer part of the circuit. The capacitor discharges through (R + R 3 ) with a time constant of (R + R 3 )C (15.0 k Ω + 3.00 k Ω)(10.0 µf) 0.180 s. The initial current I i in this discharge circuit is determined by the initial potential difference across the capacitor applied to (R + R 3 ) in series: I i ( V) C (R + R 3 ) IR (333 µa)(15.0 kω) 78 µa (R + R 3 ) (15.0 kω+3.00 kω) Thus, when the switch is opened, the current through R changes instantaneously from 333 µa (downward) to 78 µa (downward) as shown in the graph. Thereafter, it decays according to I R I i e t/(r +R 3 )C (78 µa)e t/(0.180 s) (for t > 0) (d) The charge q on the capacitor decays from Q i to Q i /5 according to (a) q Q i e t/(r +R 3 )C Q i 5 Q s) ie( t/0.180 5 e t/0.180 s ln 5 t 180 ms t (0.180 s)(ln 5) 90 ms
16 Chapter 8 Solutions 8.68 V ε e trc ε so ln ε V 1 RC t A plot of ln V versus t should be a straight line with slope 1 RC. Using the given data values: t s () V (V) ln ( ε V ) 0 6.19 0 4.87 5.55 0.109 11.1 4.93 0.8 19.4 4.34 0.355 30.8 3.7 0.509 46.6 3.09 0.695 67.3.47 0.919 10. 1.83 1.19 (a) A least-square fit to this data yields the graph to the right. Σx i 8, Σx i 1.86 10 4, Σx i y i 44, Σy i 4.03, N 8 Slope N ( Σx i y i ) Σx i N( Σx i ) Σx i ( )( Σy i ) ( ) 0.0118 Intercept Σx i ( ) Σy i ( ) Σx i N( Σx i ) Σx i ε The equation of the best fit line is: ln V ( 0.0118 )t + 0.088 ( )( Σx i y i ) ( ) 0.088 1 1 (b) Thus, the time constant is τ RC slope 0. 0118 84.7 s and the capacitance is C τ R 84.7 s 10.0 10 6 8.47 µf Ω
Chapter 8 Solutions 163 8.69 r i/6 a r r r r r r r r r r r b a i/3 i/6 i/3 i/3 i/6 i/3 i/6 i/6 i/3 i/6 i/3 b 3 junctions at the same potential another set of 3 junctions at the same potential 8.70 (a) For the first measurement, the equivalent circuit is as shown in Figure 1. a c R 1 b R ab R 1 R y + R y R y Ry Rx Ry so R y 1 R 1 (1) Figure 1 For the second measurement, the equivalent circuit is shown in Figure. a R c Thus, R ac R 1 R y + R x () Ry Ry Rx Figure Substitute (1) into () to obtain: R 1 1 R 1 + R x, or R x R 1 4 R 1 (b) If R 1 13.0 Ω and R 6.00 Ω, then R x.75 Ω The antenna is inadequately grounded since this exceeds the limit of.00 Ω. 8.71 Since the total current passes through R 3, that resistor will dissipate the most power. When that resistor is operating at its power limit of 3.0 W, the current through it is R 1 R 3 I total P R 3.0 W.00 Ω 16.0 A, or I total 4.00 A Half of this total current (.00 A) flows through each of the other two resistors, so the power dissipated in each of them is: R R 1 R R 3.00 Ω P ( 1 I total) R (.00 A) (.00 Ω) 8.00 W Thus, the total power dissipated in the entire circuit is:
164 Chapter 8 Solutions P total 3.0 W + 8.00 W + 8.00 W 48.0 W 8.7 The total resistance between points b and c is: ( 00. k Ω )( 300. kω ) R 10. kω 00. k Ω+ 300. kω The total capacitance between points d and e is: C.00 µf + 3.00 µf 5.00 µf The potential difference between point d and e in this series RC circuit at any time is: [ ] ( 10.0 V) [ 1 e 1000t 6 ] V ε 1 e trc Therefore, the charge on each capacitor between points d and e is: q 1 C 1 ( V) (.00 µf) ( 10.0 V) [ 1 e 1000t 6 ] ( 40 µc ) [ 1 e 1000t 6 ] and q C ( V) ( 3.00 µf) ( 10.0 V) [ 1 e 1000t 6 ] ( 360 µc ) [ 1 e 1000t 6 ].00 kω.00 kω 1.00 µf 3.00 µf 0 V *8.73 (a) R eq 3R I ε 3R P series ε I ε 3R 1 (b) R eq ( 1/R)+ ( 1/R)+ 1/R ( ) R 3ε 3 I R P parallel ε 3ε I R (c) Nine times more power is converted in the parallel connection.